This transfer is possible in two ways: direct transfer and using the decimal system.
First we will perform the translation through the decimal system
let\'s translate to decimal like this:
3∙162+4∙161+11∙160 = 3∙256+4∙16+11∙1 = 768+64+11 = 84310
got It: 34B16 =84310
Translate the number 84310 в octal like this:
the Integer part of the number is divided by the base of the new number system:
843 | 8 | | | |
-840 | 105 | 8 | | |
3 | -104 | 13 | 8 | |
| 1 | -8 | 1 | |
| | 5 | | |
|
the result of the conversion was:
84310 = 15138
answer: 34B16 = 15138
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
34B16 = 3 4 B = 3(=0011) 4(=0100) B(=1011) = 11010010112
answer: 34B16 = 11010010112
Fill in the number with missing zeros on the left
let\'s make a direct translation from binary to post-binary like this:
0011010010112 = 001 101 001 011 = 001(=1) 101(=5) 001(=1) 011(=3) = 15138
answer: 34B16 = 15138